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In order to prove if a relation is an equivalence relation, it needs to be show that is all of:

  • Reflexive
  • Symmetric
  • Transitive

Whilst I am familiar with this, I am unsure how to approach the following set of questions:

State and explain whether each of the following relations $R$ is an equivalence relation.

  1. The relation $R\subseteq(\mathbb{Z}\times\mathbb{Z} \ \backslash \ \{0\}) \times (\mathbb{Z}\times\mathbb{Z} \ \backslash \ \{0\}) $ is defined via $((a,b),(c,d))\in R$ if, and only if, $ad=bc$.
  2. The relation $R\subseteq(\mathbb{Z}\times\mathbb{Z}) \times (\mathbb{Z}\times\mathbb{Z} ) $ is defined via $((a,b),(c,d))\in R$ if, and only if, $ad=bc$.

The main thing I am confused about is how to deal with relations involving cartesian products of cartesian products. To prove it is reflexive, am I supposed to prove all $$((a,a),(a,a))$$ exists within the relation, if and only if $a^2=a^2$? And I'm not sure how to begin with the others. Thanks for any tips.

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  • $\begingroup$ For reflexivity, you have to show $((x,y),(x,y))\in R$ It doesn't matter what the elements of the set look like; you have to show each is related to itself. $\endgroup$
    – saulspatz
    May 12, 2019 at 15:41

3 Answers 3

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Concerning the first binary relation, not that the elements of $\mathbb Z\times(\mathbb Z\setminus\{0\})$ are pairs $(a,b)$ (with $a,b\in\mathbb Z$ and $b\neq0$, not pairs $(a,a)$. Now:

  • Reflexive: $\bigl((a,b),(a,b)\bigr)\in R$, because $ab=ab$.
  • Symmetric: It is symmetric because the definition itself is symmetric.
  • Transitive: Suppose that $\bigl((a,b),(c,d)\bigr),\bigl((c,d),(e,f)\bigr)\in R$. This means that $ad=bc$ and that $cf=de$. Do we have $\bigl((a,b),(e,f)\bigr)\in R$? In other words, is it true that $af=be$. Yes, because$$\frac ab=\frac cd=\frac ef.$$

Can you deal with the other one now? Hint: It is not an equivalence relation.

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  • $\begingroup$ This makes things much simpler. Might the second one not an equivalence relation because it is not transitive, as $\frac{0}{0}$ is not defined? $\endgroup$
    – zzxxx123
    May 12, 2019 at 15:58
  • $\begingroup$ Yes, it is not transitive, but in order to prove that you will have to find a case in which $\bigl((a,b),(c,d)\bigr),\bigl((c,d),(e,f)\bigr)\in R$, but $\bigl((a,b),(e,f)\bigr)\notin R$ $\endgroup$ May 12, 2019 at 16:00
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You can get around the "cartesian products of cartesian products" confusion by writing the relation the way you write "$=$", so $$ (a,b) R (c,d) \iff ad = bc . $$ Now the only cartesian product involves the pairs that may or may not be related. So to show that the relation is reflexive you need to show that for every pair $(a,b)$, $$ (a,b) R (a,b). $$

$xRy$ rather than $(x,y) \in R$ is in fact how equivalence relations (and order relations) are usually written in practice. Similarly, functions are often written as $y = f(x)$ rather than $(x,y) \in f$ even though, formally, a function is just a relation with certain properties.

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How to deal with cartesian products of cartesian products:

Let $\Omega := \mathbb{Z}\times\mathbb{Z}$ (or, respectively, $\mathbb{Z}\times\mathbb{Z}\setminus \{0\}$).

For reflexivity: You are supposed to prove (or disprove) $((a,b),(a,b))\in R$ for all $(a,b)\in\Omega$.

For symmetry: Prove/Disprove $((a,b),(c,d))\in R \iff ((c,d),(a,b))\in R$ for all $(a,b),(c,d)\in\Omega$.

For transitivity: Prove/Disprove $((a,b),(c,d))\in R$ and $((c,d),(e,f))\in R$ implies $((a,b),(e,f))\in R$ for all $(a,b),(c,d),(e,f)\in\Omega$.

José has already taken care of the three points above for your particular relation from 1., so I won't do that here.

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